90a,bc : a,bc = 721 

(900 + a,bc) : a,bc = 721

900 : a,bc + a,bc  : a,bc = 721

900 : a,bc + 1 = 721

900 : a,bc = 721 - 1 = 720

a,bc = 900 : 720 = 1,25

Vậy a = 1; b = 2; c = 5

=> (900 + a,bc . 0,01) : (a,bc . 0,01) = 721

=> 900 :  (a,bc . 0,01) + (a,bc . 0.01) :  (a,bc . 0,01) = 721

=> 900 : (a,bc . 0,01) + 1 = 721

=> 900 :  (a,bc . 0,01) = 720

=> a,bc . 0,01 = 1,25

=> abc = 125

Vậy a = 1 ; b = 2 và c = 5

ban j minh 200cm3=...cc nha

\(220,33-\overline{ab,cd}=\overline{abc,d}\)

\(\left(220,33-\overline{ab,cd}\right)\times100=abc\times100\)

\(22033-\overline{abcd}=\overline{abcd}\times10\)

\(22033=\overline{abcd}\times10+\overline{abcd}\)

\(22033=\overline{abcd}\times11\)

\(\overline{abcd=}22033\div11=2003\)

\(\overline{90a,bc}\div\overline{a,bc}=721\)

\(\overline{90a,bc}=721\times\overline{abc}\)

\(\overline{90abc}=721\times\overline{abc}\)

\(90000+\overline{abc}=721\times\overline{abc}\)

\(90000=721\times\overline{abc}-\overline{abc}\)

\(90000=720\times\overline{abc}\)

\(\Rightarrow\overline{abc}=\frac{90000}{720}=125\)

k mik vs

mik k lại cho

k mik vs

mik k lại cho

Ta có:  \(\hept{\begin{cases}\left|x-6\right|\ge0\forall x\\\left|y-3\right|\ge0\forall y\end{cases}}\)

Mà \(\left|x-6\right|+\left|y-3\right|=0\)

\(\Rightarrow\hept{\begin{cases}\left|x-6\right|=0\\\left|y-3\right|=0\end{cases}\Rightarrow\hept{\begin{cases}x-6=0\\y-3=0\end{cases}\Rightarrow}\hept{\begin{cases}x=6\\y=3\end{cases}}}\)

vậy....

hok tốt!!

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a, Đặt \(x=2k;y=3k\)

Ta có : \(xy=54\Rightarrow6k^2=54\Leftrightarrow k^2=9\Leftrightarrow k=\pm3\)

Với  k = 3 thì x = 6 ; y = 9 

Với k = -3 thì x = -6 ; y = -9 

b, Áp dụng tính chất dãy tỉ số bằng nhau 

\(\dfrac{x}{5}=\dfrac{y}{3}=\dfrac{x^2-y^2}{25-9}=\dfrac{4}{16}=\dfrac{1}{4}\Rightarrow x=\dfrac{5}{4};y=\dfrac{3}{4}\)

a) Đặt \(\dfrac{x}{2}=\dfrac{y}{3}=k\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=2k\\y=3k\end{matrix}\right.\)

Ta có: xy=54

\(\Leftrightarrow6k^2=54\)

\(\Leftrightarrow k^2=9\)

\(\Leftrightarrow\left[{}\begin{matrix}k=3\\k=-3\end{matrix}\right.\)

Trường hợp 1: k=3

\(\Leftrightarrow\left\{{}\begin{matrix}x=2\cdot3=6\\y=3\cdot3=9\end{matrix}\right.\)

Trường hợp 2: k=-3

\(\Leftrightarrow\left\{{}\begin{matrix}x=2\cdot\left(-3\right)=-6\\y=2\cdot\left(-3\right)=-9\end{matrix}\right.\)

b) Đặt \(\dfrac{x}{5}=\dfrac{y}{3}=k\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=5k\\y=3k\end{matrix}\right.\)

Ta có: \(x^2-y^2=4\)

\(\Leftrightarrow\left(5k\right)^2-\left(3k\right)^2=4\)

\(\Leftrightarrow25k^2-9k^2=4\)

\(\Leftrightarrow k^2=\dfrac{1}{4}\)

\(\Leftrightarrow\left[{}\begin{matrix}k=\dfrac{1}{2}\\k=-\dfrac{1}{2}\end{matrix}\right.\)

Trường hợp 1: \(k=\dfrac{1}{2}\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=3\cdot\dfrac{1}{2}=\dfrac{3}{2}\\y=5\cdot\dfrac{1}{2}=\dfrac{5}{2}\end{matrix}\right.\)

Trường hợp 2: \(k=-\dfrac{1}{2}\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=3\cdot\dfrac{-1}{2}=\dfrac{-3}{2}\\y=5\cdot\dfrac{-1}{2}=\dfrac{-5}{2}\end{matrix}\right.\)

mấy bn giúp mik ik màaaaaaaaaaaaaaaaaaaa

Ta có: \(x+5⋮x+3\)

\(\Rightarrow\left(x+3\right)+2⋮x+3\)

Vì \(x+3⋮x+3\)

\(\Rightarrow2⋮x+3\)

\(\Rightarrow x+3\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)

Ta có bảng:

Hok tốt!!

\(A=18\cdot\left(\frac{373737}{424242}+\frac{373737}{565656}+\frac{373737}{727272}\right)\)

\(=18\cdot\left(\frac{37}{42}+\frac{37}{56}+\frac{37}{72}\right)=18\cdot37\cdot\left(\frac{1}{42}+\frac{1}{56}+\frac{1}{72}\right)\)

\(=18\cdot37\cdot\left(\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+\frac{1}{8\cdot9}\right)\)

\(=18\cdot37\cdot\left(\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}\right)\)

\(=18\cdot37\cdot\left(\frac{1}{6}-\frac{1}{9}\right)=18\cdot37\cdot\left(\frac{3}{18}-\frac{2}{18}\right)\)

\(=18\cdot37\cdot\frac{1}{18}=18\cdot\frac{1}{18}\cdot37=1\cdot37=37\)